1967 AHSME Problem 40

Located inside equilateral triangle ABC is a point P such that PA=8, PB=6, and PC=10. To the nearest integer the area of triangle ABC is:

\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.